# Arguments of Roots of Complex Numbers

## Homework Statement

i) Solve the equation $z^3 = \mathbf{i}$.
(ii) Hence find the possible values for the argument of a complex number w which is such that $w^3 = \mathbf{i}(w*)^3$.

I'm stuck on part ii.

## The Attempt at a Solution

The answer to the equation in part i: e^(1/6 pi i) , e^(5/6pi i ), e^(-1/2pi i )

In the back of the book it has 6 answers, I can only get three of them. If I draw it out on the argand diagram the 3 answers I am missing would be the arguments if you carried the line on through the axis. For example I have arguments 5/12pi , pi/12, -pi/4 and am missing 3/4pi, -11/12pi and -7/12pi.

My method was to cube root both sides:
e^theta i = Z e^ -theta i
Where Z is one of the solutions to part i, by changing Z I can get 3 solutions but am missing the other three.

After seeing in the book there were six solutions I did manage to get them by letting i = e^pi/2 i and setting it equal to e^6theta i, then I just kept adding 2pi to the argument of i until I got all 6 arguments and ended up back at the beginning. However if I didn't have the answers I wouldn't have known to look for the extra three, and this method didn't involve part i) of the question (which I think it is meant to because of the 'hence').

I'd be interested to see how you all approach the problem, and an explanation of what is actually happening.

Thanks

tiny-tim
Homework Helper
Welcome to PF!

Hi I like number! Welcome to PF! (have a pi: π and try using the X2 tag just above the Reply box )

Hint: write it (w/w*)3 = i Hi I like number! Welcome to PF! (have a pi: π and try using the X2 tag just above the Reply box )

Hint: write it (w/w*)3 = i Doing that I end up with

e2itheta=3rdrt i

From part i) of the question I have all 3 values for the 3rdrt i , so i set e2itheta= to each of them one at a time , then find theta but this only gives 3 values.
If I was to plot the values I get on an argand diagram mine would be in the first and fourth quadrant, and I'm missing the 3 values from the second and third quadrants.

I hope I'm making sense, and thanks once again for the help.

I like number

tiny-tim
Homework Helper
e2itheta=3rdrt i

From part i) of the question I have all 3 values for the 3rdrt i …

ok, so you have three values for e2iθ, which is (e)2,

so e is the square-root of that, which has two values, ± of each other. ok, so you have three values for e2iθ, which is (e)2,

so e is the square-root of that, which has two values, ± of each other.  *doh*!

e= ± eπ/12i
And from this I get the value I had which was π/12 if I take the positive root, if I take the negative root then the argument will be π12 + π = 13π/12 which gives the argument -11π/12 which was one of the missing ones :D If i do this for the others I'll get them :)
Perfect, thank you tiny-tim!

I like number